Question: Simplify; express your answer in exponential form. Assume $k\neq 0, p\neq 0$. $\dfrac{{(k^{2})^{5}}}{{(k^{2}p^{-1})^{-3}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{2}}$ to the exponent ${5}$ . Now ${2 \times 5 = 10}$ , so ${(k^{2})^{5} = k^{10}}$ In the denominator, we can use the distributive property of exponents. ${(k^{2}p^{-1})^{-3} = (k^{2})^{-3}(p^{-1})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{2})^{5}}}{{(k^{2}p^{-1})^{-3}}} = \dfrac{{k^{10}}}{{k^{-6}p^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{10}}}{{k^{-6}p^{3}}} = \dfrac{{k^{10}}}{{k^{-6}}} \cdot \dfrac{{1}}{{p^{3}}} = k^{{10} - {(-6)}} \cdot p^{- {3}} = k^{16}p^{-3}$.